Solve for x: x^3+2x^2-15x>0

Answer 1

Hi Rudy O.,

First lets find the zero's (y = 0), and then we'll see what occurs outside and in between them. We need to know when y = f(x) > 0.

x³ + 2x² -15x > 0

x(x² + 2x - 15) > 0

x(x + 5)(x - 3) > 0

Our zero's (y=0) are at x = -5, 0, and 3. We need to know what occurs before and after all the zero's.

First lets see what occurs when x < -5:

(-6)³ + 2(-6)² - 15(-6) = -54, so when x < -5, y = f(x) < 0, not what we want.

Next lets see what happens between -5 < x < 0:

(-3)³ + 2(-3)² - 15(-3) = 36, so y = f(x) > 0, what we want.

Next 0 < x < 3:

2³ + 2(2)² - 15(2) = -14, y = f(x) < 0, not what we want.

Next x > 3:

4³ + 2(4)² - 15(4) = 36, y = f(x) > 0, what we want.

So for x³ + 2x² -15x > 0, [-5 < x < 0 or x > 3]

I hope this helps, Joe.